Answer:
v2=15.027m/s
Explanation:
If the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower head openings?
could be the concluding part to this question.
therefore
form equation of continuity
dv/dt=[tex]a_{1} *v_{1} =a_{2} *v_{2}[/tex]
a_{1}=[tex]\pi[/tex]*r^2
pi*(9*10^-4)^2
2.54*10^-6
tere are 20 f them 20*
a1=5.08*10^-5
a2=pi*(9*10^-3)^2
a2*v2=pi*(9*10^-3)^2*3
7.63*10^-4=5.08*10^-5*V2
v2=15.027m/s
